# 2 Simplifying the Equations@

Info

Simplifying the Equations: The Newtonian, Lagrangian, and Hamiltonian formalisms all provide the same system of equations for the motion of two particles under mutual Newtonian gravitational attraction with respect to the origin of an inertial coordinate system.

(2.1)@$\begin{split}\ddot{\mathbf{r}}_1 &= -\frac{G m_2}{|\mathbf{r}_1 - \mathbf{r}_2|^3}(\mathbf{r}_1 - \mathbf{r}_2) \\ \ddot{\mathbf{r}}_2 &= -\frac{G m_1}{|\mathbf{r}_2 - \mathbf{r}_1|^3}(\mathbf{r}_2 - \mathbf{r}_1)\end{split}$

Amazingly, this system has a closed-form solution. Obtaining it, however, is made significantly easier once several properties are known.

Description

Reframe the problem to half the amount of equations ($$\mathbf{r}$$) and parameters ($$\mu$$).

## 2.1 Relative form@

To reduce the complexity of (2.1) (as we will see), we can define the relative distance between each body as

(2.1.1)@$\mathbf{r} = \mathbf{r}_2 - \mathbf{r}_1.$

The equations of motion may be calculated directly from (2.1) since

(2.1.2)@$\begin{split}\ddot{\mathbf{r}} &= \ddot{\mathbf{r}}_2 - \ddot{\mathbf{r}}_1 \\ &= -\frac{G(m_1 + m_2)}{r^3}\mathbf{r},\end{split}$

where $$r = |\mathbf{r}|$$ is the distance between the two particles.

Important

Doing this is beneficial because it exploits a symmetry in the full system of equations, essentially halving the number of equations to be solved in order to obtain any information.

### 2.1.1 The gravitational parameter@

Equation (2.1.2) will actually be the one that we will use to analyze the 2-body problem. As such, the quantity $$G(m_1 + m_2)$$ will be appearing a lot, so it is convenient to define a new symbol. This symbol is called the gravitational parameter and is denoted

(2.1.3)@$\mu := G(m_1 + m_2).$

This is convenient because we now only need to write one symbol, but more importantly, the motion only depends on one parameter — the total mass $$m_1 + m_2$$, not necessarily what $$m_1$$ and $$m_2$$ are individually.

• Example: Suppose $$\mu = 1$$. The motion $$\mathbf{r}$$ will be the same regardless whether $$(Gm_1, Gm_2) = (0.5, 0.5)$$ or $$(Gm_1, Gm_2) = (0.1, 0.9)$$.

Using the gravitational parameter, equation (2.1.2) is now written

(2.1.4)@$\ddot{\mathbf{r}} = -\frac{\mu}{r^3}\mathbf{r}.$

Important

We have decreased the complexity of studying the equations of motion since they now depend on only a single parameter ($$\mu$$) as opposed to two ($$Gm_1$$ and $$Gm_2$$).

### 2.1.2 Linear momentum@

The two bodies are isolated — that is, the center of mass satisfies

$\mathbf{r}_{cm}(t) = \mathbf{r}_{cm}(0) + \dot{\mathbf{r}}_{cm}(0) t,$

since there is no external forcing to the system. As such, the center of mass moves in a straight line (or not at all) in accordance with Newton’s 1st law. Further, we have explicitly that

(2.1.5)@$\mathbf{r}_1 = \mathbf{r}_{cm} - \frac{m_2}{m_1 + m_2}\mathbf{r} \qquad \text{and} \qquad \mathbf{r}_2 = \mathbf{r}_{cm} + \frac{m_1}{m_1 + m_2}\mathbf{r}.$

You can validate this for yourself!

Hint

For any $$\mathbf{x}_1, \mathbf{x}_2 \in \mathbb{R}^3$$, the center of mass for a system of two particles with respect to an inertial coordinate system is

$\mathbf{x}_{cm} := \frac{m_1 \mathbf{x}_1 + m_2 \mathbf{x}_2}{m_1 + m_2},$

where $$m_1$$ and $$m_2$$ are the masses of each body located (instantaneously) at $$\mathbf{x}_1$$ and $$\mathbf{x}_2$$.

Important

(2.1.1) provides direct information of both bodies (relative to each other) through (2.1.2) and enables the inertial positions of both bodies to be calculated from (2.1.5) rather than directly solved from (2.1).

This solidifies the idea that only half of the amount of equations have to be solved with (2.1.2), but now all information about both bodies is known.

## 2.2 The reduced mass@

The system (2.1.2) is writable as

$\underbrace{\frac{m_1 m_2}{m_1 + m_2}}_{\mu^*}\ddot{\mathbf{r}} = \underbrace{-\frac{G m_1 m_2}{r^3}\mathbf{r}}_{-\nabla V},$

which is in the form of Newton’s 2nd law for a single particle of mass $$\mu^*$$ being tracked in an inertial coordinate system with position $$\mathbf{r}$$ under the influence of a potential $$V$$. More clearly, we can write

$\mu^* \ddot{\mathbf{r}} = -\frac{G(m_1 + m_2)\mu^*}{r^3}\mathbf{r}$

The 2-body system (2.1.2) can therefore be treated like it describes a single particle1. The quantity $$\mu^*$$ that makes this purely mathematical simplification possible is called the reduced mass.

Important

The trajectory of a single body of mass $$\mu^*$$ under the influence of the potential from a static body of mass $$m_1 + m_2$$ is the SAME trajectory experienced by the relative motion of two bodies under mutual Newtonian gravitational attraction! This trajectory for both cases is $$\mathbf{r}$$.

### 2.2.1 Quick facts about the reduced mass system@

Fact: $$\mu^* < m_1 + m_2$$ for any $$m_1, m_2 > 0$$.

Proof

Suppose the opposite. Then

$\begin{split}(m_1 + m_2)^2 &= m_1^2 + 2m_1 m_2 + m_2^2 \\ &< m_1 m_2\end{split}$

but this means

$\begin{split}0 &< m_1^2 + m_2^2 \\ &< -m_1 m_2 \\ &< 0\end{split}$

a contradiction. $$\quad\square$$

• This means that it’s fair to imagine the smaller body simply flying around the larger body.

Fact: $$\mu^* \to m_2$$ when $$m_1 \gg m_2.$$

Proof

Take $$m_1 = k m_2$$ so that

$\frac{\mu^*}{m_2} = \frac{k}{k + 1}$

and let $$k \to \infty$$. $$\quad\square$$

• This means that when the reduced mass is very small, then this system actually resembles the relative system. That is, $$m_1$$ is nearly sitting at the origin and $$m_2$$ is flying around it.

Important

In our solar system, the gravitational parameter can be easily regarded as

$\mu \approx G m_1.$

This is the case where $$m_2$$ represents spacecraft/moons/comets such that $$m_1 \gg m_2.$$

The effective statement of taking $$\mu$$ this way is that the central body of mass $$m_1$$ moves in a straight line (or not at all) relative to the inertial frame but the motion of the smaller body of mass $$m_2$$ is still affected by the presence of the central body.

Warning

Do not confuse the 2-body gravitational parameter $$\mu$$ with the 3-body mass parameter $$\mu$$.

These quantities, though sharing the same symbol, are completely different.

1

The Two-Body Problem - UCSB Physics